LeetCode 之 Partition List — C++ 實現

Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

給定一個鏈表和一個值 x，將其分區使 x 後面的節點值全部大於等於 x。

必須保證原來值的相對順序。

分析：

    從表頭開始掃描，小於 x 的值跳過，遇到大於等於 x 的值則向後查找小於 x 的值，然後插入到大於等於 x 值的前面。

class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        if(!head)
        {
            return NULL;
        }
        
        ListNode *tempHead = new ListNode(0);
        tempHead->next = head;
        ListNode *pre = tempHead, *p = head, *pr = NULL, *cur = NULL;
        
        while(p)
        {
           if(p->val < x) //val < x, 繼續向後搜索
            {
                pre = pre->next;
                p = p->next;
            }
            else //val > x
            {
                pr = pre->next; //pr記錄 val < x 的前一個節點
                p = p->next;
                while(p && p->val >= x)
                {
                    pr = pr->next;
                    p = p->next;
                }
                if(p)//找到 val < x 的節點
                {
                    pr->next = p->next;
                    p->next = pre->next;
                    pre->next = p;
                    
                    pre = pre->next;
                    p = pre->next;
                }
            }
        }
        
        head = tempHead->next;
        delete tempHead;
        
        return head;
    }
};